On Maximizing Coverage in Gaussian Relay Networks

arXiv:0802.2360v1 [cs.IT] 17 Feb 2008

On Maximizing Coverage in Gaussian Relay Networks Vaneet Aggarwal

Amir Bennatan

A. Robert Calderbank

Department of Electrical Engineering Princeton University Email: [email protected]

Program for Applied and Computational Mathematics Princeton University Email: [email protected]

Department of Electrical Engineering Princeton University Email: [email protected]

Abstract—Results for Gaussian relay channels typically focus on maximizing transmission rates for given locations of the source, relay and destination. We introduce an alternative perspective, where the objective is maximizing coverage for a given rate. The new objective captures the problem of how to deploy relays to provide a given level of service to a particular geographic area, where the relay locations become a design parameter that can be optimized. We evaluate the decode and forward (DF) and compress and forward (CF) strategies for the relay channel with respect to the new objective of maximizing coverage. When the objective is maximizing rate, different locations of the destination favor different strategies. When the objective is coverage for a given rate, and the relay is able to decode, DF is uniformly superior in that it provides coverage at any point served by CF. When the channel model is modified to include random fading, we show that the monotone ordering of coverage regions is not always maintained. While the coverage provided by DF is sensitive to changes in the location of the relay and the path loss exponent, CF exhibits a more graceful degradation with respect to such changes. The techniques used to approximate coverage regions are new and may be of independent interest.

rate). Consider a source, relay and destination all at equal distances (say 1) from one another (on the vertices of an equilateral triangle), as in Fig. 1. Assume equal power constraints (P1 = P2 = 1). Assume the requisite rate is R = 1 bits/channel use. Then DF can achieve a maximum rate of 1 bit/channel use at the destination, yet CF can do better: 1.17. In this paper (Theorem 1, Sec. III) we assert that whenever the relay can decode, DF provides superior performance to that of CF. This example would appear to contradict that assertion. However, with the performance measure of our paper (coverage), the advantage of CF does not matter. This is because we are not concerned with the maximum achievable rate, as long as it is greater than R.

I. I NTRODUCTION Relay channels have recently attracted significant attention as a model for ad-hoc networks [7]. These channels model problems where one or more relays help a pair of terminals communicate. The general channel model was first considered by van der Meulen [1], [2], [3] and further studied in a ground breaking work by Cover and El Gamal [4]. Although the capacity region for the channel is still unknown, the results of [4] include two achievable coding strategies which were subsequently named decode-and-forward (DF) and compressand-forward (CF). The Gaussian relay channel was examined by Kramer et al. [5] and Høst-Madsen and Zhang [8]. A distinctive property of the existing results in the literature is that they all consider the classic information-theoretic perspective of maximizing the achievable rate for given locations of the source, relay and destination nodes. However, in many cases of practical interest, the design problem at hand is to maximize coverage for a fixed desired transmission rate. This is the focus of our paper. The following simple example, illustrates the difference between the performance measure considered in this paper (maximizing coverage), and the classic measure (maximizing

Fig. 1. Locations of the source, the relay and the destination in the above example.

In their seminal work, Kramer et al. [5][Sec. VII.B] extended the classic treatment of the relay problem by considering the location of the relay as a design parameter. That is, they considered the effect of relocating the relay on the achievable rates at the destination. This effect is heavily dependent on the destination’s location. While a destination at some locations may benefit from relocating the relay, a destination at other locations will suffer. However, in many cases of practical interest, the location of the relay is determined at a time when the destination’s location is unknown. The destination is typically a mobile station, while the relay is often a fixed terminal, whose location is determined once, at the time that the network is designed. Nevertheless, while the location of the destination may be unknown at the time of network design, the target transmission rate is typically known. Thus, the effect of changes in the

relay (change of location or communication strategy) on the coverage region can be evaluated. Our analysis further extends the discussion of [5] in the following ways, 1) The discussion of [5] is limited to a single destination, at a fixed distance of 1 (normalized distance metric) from the source, and to a relay that is located on the line segment connecting the source to the destination. Our discussion is completely general. 2) In [5][Remark 31], the authors analyze the performance of DF and CF in the limit when the relay is either close to the source or close to the destination. They provide numerical data when the relay is intermediate these two extremes. In this paper, we provide rigorous analysis for all possible locations of the relay. 3) The results of [5] provide the following intuition: Whenever the channel from the source to the relay is strong enough to enable the relay to decode the source’s message, DF renders superior performance to that of CF. However, regardless of how strong (or weak) this channel may be, there is always some rate (however low) that the channel can support (under the channel model both paper share). Strictly speaking, with the formulation of [5], the relay can always decode, and thus the intuition cannot be stated formally. The introduction of a target transmission rate, in this paper, enables us to formalize and prove the intuition. We begin in Sec. II by providing some background on the channel model and achievable strategies for it. We also formally define the concept of coverage. In Sec. III we compare the coverage regions of the CF and DF achievable strategies, for different locations of the relay. In this comparison, we assume that the relay’s location is the same with both strategies. We extend the comparison in Sec IV, and allow each strategy its own preferred relay location. In Sec. V we provide bounds on the area (measured in normalized units of area) of the coverage region of DF, as a function of the relay’s location. The discussion in this paper mostly focuses on a full-duplex non-fading channel model. In Sec. VI we briefly discuss additional channel models. Sec. VII concludes the paper. Throughout the paper, proofs are deferred to the appendix. II. BACKGROUND AND D EFINITIONS A. The Channel Model In this section, we introduce a simple relay channel model, which will be our focus throughout most of the paper. In this model, nodes are assumed to be full-duplex. This means that a node can receive and transmit simultaneously. Furthermore, the signal attenuation between any two points assumed to be a deterministic function of the distance between the two. In Sec. VI we will consider additional channel models (random fading and half-duplex). Our model is depicted in Fig. 2. The channel consist of three nodes: a source (node 1), a relay (node 2) and a

destination (node 3). We consider a two-dimensional domain for our three-node network. This means that the source, relay and destination are associated with two-dimensional location vectors a1 , a2 and a3 , respectively. For simplicity, and without loss of generality, we may assume that a1 = (0, 0), and a2 = (d, 0) where d > 0 is the distance between the source and the relay. The relations between the channel outputs and Destination 3

d13 d23 P1 1

2 P2

d12

Source

Relay

Fig. 2.

Single Relay Network

inputs are a function of the distances between the various nodes. We let dkl , k, l = 1, 2, 3 denote the distances between nodes k and l. With this notation, d12 = ka2 k = d, d13 = ka3 k and d23 = ka3 − a2 k where k · k denotes the Euclidean norm. See Fig. 2. The channel equations are now given by, y2 [i] =

1 α/2 x1 [i] d12

+ z1 [i]

y3 [i] =

1 α/2 x1 [i] d13

+

1 α/2 x2 [i] d23

+ z2 [i]

where, x1 [i] and x2 [i] are the signals transmitted from the source and relay, respectively, at time i. These signals are subject to average power constraints P1 and P2 , respectively. y2 [i] and y3 [i] denote the observed signals at the relay and destination, respectively. z1 [i] and z2 [i] are mutually independent i.i.d circularly-symmertic complex Gaussian noise with variance 1. α ≥ 2 is the path loss exponent. B. Codes and Achievable Strategies A code for the relay channel of rate R and block-length n, consists of a pair (C, {fi }ni=1 ). C is a set of 2nR codewords of length n. The source encoder constructs its signal by selecting a codeword x1 ∈ C. At time i, the source sends index x1 [i] and the relay sends x2 [i] using fi based on its past observation. That is, x2 [i] = fi (y2 [i − 1], ..., y2 [1]). A relay transmission scheme S is formally a collection of relay codes. Definition 1: Given locations a2 = (d, 0) and a3 of the relay and the destination respectively, a rate R is defined to be achievable by a scheme S (equivalently: S supports R) if for any  > 0, there exist (C, {fi }ni=1 ) ∈ S such that the rate of C is at least R, and the probability of error, under maximum-likelihood decoding is at most . We define the capacity of S at relay location a2 = (d, 0) and destination location a3 as, CS (d, a3 ) = sup{R : S supports R}

Cover and El Gamal [4] introduced two achievable coding strategies which were subsequently named decode-andforward (DF) and compress-and-forward (CF). With DF, the relay decodes the message transmitted by the source. It then cooperates with the source to transmit the message to the destination. With CF, the relay considers the observed signal from the source as a raw signal, compresses it, and transmits it to the destination. The destination then combines this observation with its own observation, and uses both to decode the source’s message. An important distinction between the two strategies, is that with DF, the relay attempts to decode the source’s message, while with CF it does not. A comprehensive description of the strategies is available e.g. in [4], [5], [8]. The achievable rates with both schemes, for our channel model, were computed in [5], [8], and are provided in Appendix A. Following their example, we confine our attention to CF when the random variables used in the generation of the codebooks are Gaussian. C. Coverage We are now ready to formally define the concept of coverage. Definition 2: Let R > 0 be a desired transmission rate. For a fixed distance d between the source and the relay, and a fixed transmission scheme S, we define the coverage region as, ∆ GS (d) = {a3 : CS (d, a3 ) ≥ R} The concept of coverage is closely related to outage - we fix a target rate, and seek to maximize the geographic region, outside which an outage occurs.

The proof is provided in Appendix B. We now interpret the results of Theorem 1. The condition d ≤ dc determines whether the relay is still able to decode the information that is transmitted to it by the source. Whenever the relay is able to decode, we see that DF is the method of choice. DF is uniformly superior in transmission to any point, in the sense that it provides coverage at any point served by CF. However, if the relay is not able to decode the data, then DF cannot be applied, and GDF (d) = ∅. For such values of d, the best approach is CF, where the relay and destination combine their channel observations and perform collaborative decoding. From a design perspective, while for d ≤ dc DF enjoys a larger coverage region than CF, it suffers from a sharp drop in performance when d crosses dc . In practical settings, when the path loss exponent α is not known (and consequently dc , as defined by (1), is not known), this may become an important disadvantage. CF, in comparison, enjoys a more graceful degradation with respect to changes in d. The above results can equivalently be stated as follows: Consider the combined transmission strategy CF ∨ DF, under which the various terminals are free to select the best of CF and DF. Theorem 1 implies that,  GDF (d), d ≤ dc ; GCF∨DF (d) = GCF (d), d > dc ; Figs. 3 and 4 present numerical examples of the regions considered in Theorem 1. In Fig. 3, d ≤ dc and in Fig. 4, d > dc . In both figures we also compare the coverage regions with a region computed according to the upper-bound (UB) as provided by [5]. P1=10, P2=10, alpha=3.52,R=3 1.5 UB DF CF NR

III. C OMPARISON OF CF AND DF In this section, we consider the coverage region when using the DF and CF approaches, for a fixed, given location of the relay a2 . For reference, we also consider the no-relay (NR) coverage region, i.e, the coverage region when the relay is not used. We would expect different locations of the destination a3 to favor different schemes. That is, for a fixed rate R, some locations would be covered only by CF, others only by DF and yet others by both. Surprisingly, however, the following result indicates a monotonic ordering of the coverage regions. Theorem 1: Assume the channel model of Sec. II and let R > 0. Let dc be defined as follows, 1/α  P1 (1) dc = 2R − 1 1) If d ≤ dc , then GDF (d) ⊇ GCF (d) ⊇ GNR (d) 2) If d > dc , then GCF (d) ⊇ GNR (d) ⊇ GDF (d) = ∅

1

0.5

0 Source

Relay

−0.5

−1

−1.5 −1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

Fig. 3. Coverage regions when d ≤ dc , P1 = 10, P2 = 10, α = 3.52, R = 3

IV. C OMPARISON WITH U NEQUAL R ELAY P LACEMENT The analysis of Sec. III focused on coverage regions for a given distance d between the source and the relay. However, different schemes may favor different locations of the relay. Thus, in this section, we provide some interesting results that

P1=10, P2=10, alpha=3.52,R=4 1.5

Theorem 3: Assume P1 = P2 and α = 2. Then for all 0 < d ≤ dc , p p 1 − a/2 2 ·d (2) π λγ · d2 ≤ |GDF (d)| ≤ π λγ · √ 1−a

UB CF NR

1

0.5

0

Source

Relay

where,

−0.5

s ρ =

−1

−1.5 −1

−0.5

0

0.5

1

1.5

Fig. 4. Coverage regions when d > dc , P1 = 10, P2 = 10, α = 3.52, R = 4

λ

=

γ

=

a

=



1−

d dc

2

1 1 + + 4 1−ρ 2 1 − 1−ρ 4 γ 1− λ

, p

(3)

2(1 + ρ) 1 − ρ2

(4) (5) (6)

consider CF and DF when they are allowed different locations of the relay. Theorem 2: Let R > 0. Let dc be defined as in Theorem 1. Then the following assertions holds: 1) Assume α = 2. Then there exists a non-negative positive fraction 0 < γ < 1/9, independent of R, such that if P2 > γ · P1 , then for all d > dc ,

Remark 1: 1) Recall, from Theorem 3, that when d > dc , GDF (d) = ∅ and thus |GDF (d)| = 0. 2) Observe that ρ is is a function of d, and implicitly a function of of R and P1 through its dependence on dc , which was defined by (1). Consequently, the other parameters are functions of these values too.

GCF (d) ⊆ GDF (dc )

The proof of this theorem is achieved by bounding the coverage region from within whose boundary is √ by an ellipse √ given by the points ( d2 + λd cos(θ), γd sin(θ)), and from outside by aq conic whose boundary is given ( d2 + q by the points√ √ √ λd cos(θ) 1 − a sin2 (θ), γd sin(θ) 1 − a sin2 (θ)/ 1 − a). These two bounding shapes, for the case of P1 = P2 = 10, R = 1, are plotted in Fig. 5, along with the true region (computed numerically). The details of the proof are provided in Appendix D.

2) There exists β(R) > 0 such that if P2 < β(R) · P1 , then there exists d0 such that, [ GCF (d0 ) * GDF (d) d>0

The proof of this theorem is provided in Appendix C Theorem 2, combined with Theorem 1, provide us with insight into the choice between strategies DF and CF when the relay location may be optimized independently for each scheme. In the case of α = 2, when the relay power is sufficiently large, DF is the method of choice regardless of the desired R. If we choose to place the relay at location d ≤ dc , then Theorem 1 tells us that GCF (d) ⊆ GDF (d), and thus DF is superior. If d > dc , then Theorem 2 tells us that GCF (d) ⊆ GDF (dc ), and thus switching from CF and DF and repositioning the relay at d = dc would render superior performance. Regardless of α, if the power at the relay is sufficiently low, Theorem 2 tells us that there are locations and rates that may only be supported by CF. These cannot be supported by DF, regardless of where we place the relay.

5 4 Actual Area Lower Bound Upper Bound

3 2 1

Relay dc

0 Source −1 −2 −3 −4 −5 −4

Fig. 5.

−2

0

2

4

6

8

Bounds for GDF , P1 = P2 = 10, R = 1, α = 2

V. B OUNDS ON THE DF C OVERAGE A REA So far, our discussion has focused on a comparison of coverage regions with DF and CF. However, a natural question that arises is what the area (in normalized units of area) of the coverage region is, and the effect of the distance d on it. In this section we partially answer this question for the DF coverage region in a few specific cases. We let |GDF (d)| denote this area, and begin with the following theorem.

Fig. 6 presents the bounds in (2), along with the true area (computed numerically) corresponding to the same parameters as Fig. 5. Examining this figure, we see that the bounds are very tight. √By (2), the ratio between the two bounds is (1 − √ a/2)/( 1 − a), which increases with d from 1 to 2 √ 6+2 √ ≈ 1.02216. Thus, the gap between the two bounds 7(5+4 2)

never exceeds 2.22%.

130

65

Actual Area Lower Bound Upper Bound

120

64

62

100

Area

Area

Actual Area Lower Bound

63

110

61 60

90 59

80 70

58 57

0

0.5

1

1.5

2

2.5

3

3.5

d12

56

0

0.5

1

1.5

2

2.5

3

3.5

d12

Fig. 6.

Bounds for GDF , P1 = P2 = 10, R = 1, α = 2

Fig. 7.

Examining Fig. 6, we may observe that |GDF (d)| approaches its maximum as d → 0. To see this, observe that both bounds coincide as d → 0, and the upper bound decreases with d. Thus, from the point of view of maximizing coverage, the relay should optimally be placed as close to the base station as possible. While this results holds for α = 2, it is not generally true, as the discussion below will show. The following theorem extends the lower bound of (2) to the case of α = 4. Theorem 4: Assume P1 = P2 and α = 4. Then for all 0 < d ≤ dc , p (7) |GDF (d)| ≥ π λγ · d2

Lower bound for GDF , P1 = P2 = 100, R = 1, α = 4

bound is greater than its value at d = 0. Changing variables from d to ρ as in the proof of Lemma 1 (Appendix F), we obtain that at ρ = .93, p |GDF |ρ=.93 ≥ π λγd2ρ=.93 = 2.0441..πd2c >

2πd2c

(a)

|GDF (d = 0)|

=

where (a) follows as in the proof of Lemma 1. VI. A DDITIONAL C HANNEL M ODELS

where, s ρ

r γ

=

 1−

=

d dc

A. Half Duplex Relay

4

2 − 1/4 1−ρ

(8) (9)

and λ is the largest real-valued solution of the equation,  4  2 1 2 1 x− − x− − 4 1−ρ 4   4 1 1 x− − =0 (10) 2 1−ρ 4 1 − ρ2 This solution can be found analytically by applying the Ferrari method, see e.g. [6][page 32]. The proof of this theorem follows along the lines of the proof of the lower bound in Theorem 3, namely by bounding the coverage area from within by an ellipse. The details of the proof are provided in Appendix E. Fig. 7 compares the lower bound of Theorem 4 with the true area (computed numerically), when P1 = P2 = 100 and R = 1. Lemma 1: The lower bound (7) becomes tight as d → 0. The proof of this lemma is provided in Appendix F. Recall that with α = 2, we proved that the maximum coverage area was achieved when d → 0. Lemma 1 enables us to show that this is not the case with α = 4. To prove this, all we need to do is find a nonzero value of d at which the lower

The half-duplex relay model is characterized by a relay that cannot transmit and receive at the same time. A comprehensive discussion of this case is provided in [9], [5] and [10]. In this setting, we assume that there exists a positive fraction t ∈ [0, 1] such that the relay is listening (receiving) during a proportion t of the time, and transmitting during a proportion 1 − t of the time. For this setting, we have the following theorem, Theorem 5: Assume the half-duplex channel model and let R > 0. The following results hold. 1) If t = 1/2 and d < d0c , where d0c =



P1 22R − 1

1/α (11)

then GDF (d) ⊇ GCF (d) ⊇ GNR (d) 2) If d > dc , then for all t ∈ [0, 1] GCF (d) ⊇ GNR (d) ⊇ GDF (d) The proof is provided in Appendix G. This theorem is weaker than Theorem 1, because d0c < dc . Furthermore, note that we have confined our attention to the case that t = 1/2.

B. Random Fading Models In this section, we modify the channel model of Sec. II to introduce some random fading. The new channel equations are now given by,

1.5 UB DF NR CF

1

0.5

y2 [i] =

h12 jϕ12 [i] x1 [i] α/2 e d12

+ z1 [i]

y3 [i] =

h13 jϕ13 [i] x1 [i] α/2 e d13

+

h23 jϕ23 [i] x2 [i] α/2 e d23

0 Source

+ z2 [i]

Relay

−0.5

We consider two fading models, 1) Phase fading hkl = 1 for all {k,l}. ϕkl [i] are uniformly distributed over [0, 2π), and are jointly independent of one other, of the transmitted signals and the noise. Their time realizations are also independent. 2) Rayleigh fading. ϕkl [i] is defined as in the phase fading case. hkl are Rayleigh distributed with parameter 1, independent, and remain fixed for the duration of the transmission. In all cases, we assume that the realizations of the random variables hkl and ϕkl [i] are known to the receivers but not to the transmitters. We begin by considering the phase-fading model. In Appendix H-A we will show that Theorem 1 carries over directly to this case. Furthermore, Kramer et al. [5] have computed an upper bound (UB) on the capacity in the phase-fading model. In Appendix H-B we prove the following lemma. Lemma 2: Assume d ≤ dc where dc is given by (1). Then GUB (d) = GDF (d). We now proceed to provide some interesting observation for the Rayleigh fading model. This channel is no longer an ergodic channel, and thus we redefine the coverage region in terms of outage probabilities ([7], [5]). Definition 3: Let R > 0 be a desired transmission rate and 0 <  < 1 be the maximum tolerable outage probability. For a fixed distance d between the source and the relay, and a fixed transmission scheme S, we define the coverage region as, ∆

GS (d) = {a3 : P r[CS (d, a3 ) ≥ R] ≥ 1 − } We define dˆc in a manner analogous to dc of Theorem 1, dˆc =



−P1 ln(1 − ) 2R − 1

1/α

dˆc is the distance after which the relay cannot decode with probability greater than 1 − . At relay locations satisfying d > dˆc , DF cannot be applied. Expressions for the achievable rates with CF and DF in this case are available in [5]. In Fig. 8 we have plotted the coverage regions for the various schemes in the Rayleigh fading model. Interestingly, the monotonic ordering of GCF (d) and GDF (d), which was observed in Theorem 1 for the simple (non-fading) model, is not maintained. Neither of the two regions contains the other.

−1

−1.5 −1.5

−1

−0.5

0

0.5

1

1.5

2

Fig. 8. Coverage regions with Rayleigh Fading and d < dˆc , P1 = 3, P2 = 0.3, α = 3.52, R = 1,  = 0.35, d = 0.9

VII. C ONCLUSION In this paper, we have introduced a new perspective on the relay channel, switching from maximizing rate at a fixed given destination location, to maximizing coverage for a given rate. This perspective opens up an array of new possibilities for research. Our main contributions are first, the formulation of the problem in Sec. II. In Theorem 1, we have obtained the surprising result that for any given placement of the relay, one of the two common strategies (CF and DF) is uniformly optimal, and thus a relay that switches between the two is not required. Theorem 2 extends the comparison for the case of α = 2 by allowing each of the strategies its own preferred relay location. In this case, it is interesting that the results depend on the power constraint on the relay. Theorems 1 and 2 imply that while DF often provides a larger coverage region than CF, it is also more sensitive to changes in the location of the relay and the path loss exponent α. In contrast, CF is more robust and provides a more graceful degradation with respect to such changes. A natural question that arises, when considering coverage regions, is the numeric area of the region, as a function of the distance between the source and the relay. In Sec. V we have provided bounds in two special cases. In Sec.VI, we have have extended our discussion to halfduplex and random-fading channel models. In particular, our results for fading channels indicate that the monotonic ordering of the DF and CF coverage regions no longer holds. A further study of the effect of relocating the relay, in such settings, is of great practical interest. In this paper, we have focused on a one-relay setting. An interesting extension would consider two or more relays. In this case, the number of degrees of freedom grows substantially, as not only distance to the relays but also the angles between the line segments connecting them to the source, may be optimized. An analysis of the benefit from cooperation between the relays, is also of great interest.

The following lemma examines this expression at x = xDF (θ).

A PPENDIX A G ENERAL E XPRESSIONS AND N OTATION A. DF and CF Achievable Rates The achievable rates, with DF and CF, that were computed by [5], [8],    P1 2 CDF = max min log 1 + α (1 − ρ ) , 0≤ρ≤1 d12 !) √ P1 P2 2ρ P1 P2 log 1 + α + α + α/2 α/2 (12) d13 d23 d13 d23 ! P1 P1 CCF = log 1 + α + (13) ˆ2 ) dα d12 (1 + N 13 ˆ2 is given by, where N α α ˆ2 = P1 (1/d12 + 1/d13 ) + 1 N α P2 /d23

(14)

B. Switch to Polar Coordinates We frequently use the polar coordinates x and θ to parameterize the destination’s location, i.e. we assume the destination is placed at, a3 = (x cos θ, x sin θ)

(15)

Letting d denote the distance between the source and the relay, the distances d12 , d13 and d23 (see Sec. II-A) now satisfy, d12 = d, d13 = x, and d223 = d2 + x2 − 2dx cos θ. A PPENDIX B P ROOF OF T HEOREM 1 We begin by noting that for all d > 0, the result GCF (d) ⊇ GNR (d) is straightforward from (13) and from the observation that,   P1 (16) CNR = log 1 + α d13 We now prove Part 2 of the theorem, which is the easier of the two parts. Whenever d > dc , by (1),     P1 P1 R > log 1 + α = max log 1 + α (1 − ρ2 ) > CDF d d ρ∈[0,1] where the last inequality was obtained by (12), recalling that d12 = d. This is true regardless of the destination location a3 , and thus DF cannot support a rate of R anywhere and GDF (d) = φ. Before proceeding to the proof of Part 1 of the theorem, we introduce the following notation. For a given scheme S (S being DF or CF), and a given θ, we define xS (θ) as the maximum x such that a destination with polar coordinates (x, θ) (see Appendix A) is contained in GS (d). The proof now focuses on showing that xDF (θ) ≥ xCF (θ) for all θ. By (12), we may obtain a lower bound on CDF by restricting the maximization to ρ = 0. Thus,      P1 P2 P1 CDF ≥ min log 1 + α , log 1 + α + α (17) d12 d13 d23

Lemma 3: For all θ, consider the point a3 whose polar coordinates are given by (θ, xDF (θ)). Then the following holds at a3 , P1 P2 P1 > α + α (18) dα d d 12 13 23 Proof: Assume, by contradiction, that (18) does not hold. Letting x = xDF (θ), we will now show that we may increase x and preserve CDF ≥ R, contradicting the maximality of xDF (θ). Consider (12). Whenever (18) does not hold, we have, for all ρ,     P1 P1 log 1 + α (1 − ρ2 ) ≤ log 1 + α d12 d12 ! √ P1 P2 2ρ P1 P2 < log 1 + α + α + α/2 α/2 d13 d23 d13 d23 Thus, the minimization in (12) is achieved by the first term. Changing x does not affect this term. Increasing x affects d13 and d23 and thus reduces the second term. However, by a continuity argument, we may increase x slightly without altering the invalidity of (12). Thus, the minimization in (12) would still be achieved by the first term, and CDF will not change. Therefore, if CDF ≥ R at x = xDF (θ), it will remain so after we increase x a little, producing the desired contradiction. By Lemma 3, (17) implies that at x = xDF (θ),   P1 P2 CDF ≥ log 1 + α + α (19) d13 d23 > CCF (20) The last inequality may be obtained by simple arithmetic using (13) and (14). By a continuity argument, CDF = R at (θ, xDF (θ)), and thus the above inequality implies that CCF < R at this point. We now argue that this result implies that GDF (d) ⊇ GCF (d). This would follow if we could show that CCF decreases in the range x > xDF (θ), for all θ. In the range x > d, d13 , d23 can easily be shown to be ˆ2 increases (see (14)), increasing functions. This implies that N and consequently CCF decreases (see (13)). Thus, our desired result would now follow if we could prove that xDF (θ) > d for all θ. To see this, observe that by (16) and (1), GNR is a sphere with radius dc . We have shown that GNR ⊂ GCF . Thus, if CCF < R at (θ, xDF (θ)) (as we have shown above), then xDF (θ) ≥ dc ≥ d (the last inequality being one of the conditions of this part of the theorem). This is precisely the result we sought, thus completing the proof of Part 1 of the theorem. A PPENDIX C P ROOF OF T HEOREM 2 Before beginning with the proof, we argue that we may assume, without loss of generality, that P1 (the power of

the source) is 1. To see this, observe that if P1 6= 1, we may replace P1 and P2 by Pˆ1 = 1 and Pˆ2 = P2 /P1 . This substitution preserves the ratio Pˆ1 /Pˆ2 = P1 /P2 , and has the effect of scaling the DF and CF coverage regions. That is, if we relocate the relay and the destination from any locations 1/α 1/α a2 and a3 (respectively) to 1/P1 · a2 and 1/P1 · a3 , the achievable rates CCF and CDF would remain unchanged. Under these assumptions, given that the relay is placed at (d, 0), and using the notation of Sec. A, the expressions for the DF and CF achievable rates (12) and (13) may be rewritten as,    1 CDF = max min log 1 + α (1 − ρ2 ) , 0≤ρ≤1 d !) √ P2 2ρ P2 1 log 1 + α + α + (21) α/2 x d23 dα/2 d23 ! 1 1 (22) + CCF = log 1 + ˆ 2 ) xα dα (1 + N and (14) by, α α ˆ2 = 1/d + 1/x + 1 N α P2 /d23

ˆ2 ), which Part 2 follows by bounding the term d2 (1 + N appears in (22), ˆ2 ) d2 (1 + N

d2 (1/d2 + 1/xα + 1 + P2 /d223 ) P2 /d223 2 2 2 d d23 (1/d + P2 /d223 ) P2 d223 + P2 d2 P2

= ≥ =

+ Our aim now is to show that wherever CCF (x, θ) ≥ R, also C˜DF (x, θ) ≥ R. In the sequel, we use the shorthand notation ∆ s = cos θ. Lemma 5: Let θ be fixed, + 1) If s ≤ 0, then for all x > 0, C˜DF (x, θ) ≥ CCF (x, θ). 2) Assume s > 0 and let x1 be given by, √ P2 + 1 1 √ √ · dc (26) x1 = s P2 + 1 − P2 Then the following holds, + CCF (x, θ) ≤ C˜DF (min(x, x1 ), θ)

(23)

A. Proof of Part 1 of Theorem 2 In this section, we assume that α = 2. Rather than focus on CDF and CCF as in (21) and (22), the following lemma allows us to revert to two alternative functions. Lemma 4: Let CDF (x, θ, d) and CCF (x, θ, d) denote the achievable rates when the relay is placed at (d, 0) and the destination location is derived from x and θ. Let, ! 1 P2 ∆ ˜ CDF (x, θ) = log 1 + 2 + 2 (24) x d23,c   1 P ∆ 2 +  CCF (x, θ) = log 1 + 2 + x inf (P2 d2 + d223 ) d>dc

(25) where d23,c is the distance from a relay placed at (dc , 0) to the destination, and d23 is the same for a relay placed at (d, 0). Then the following holds: 1) C˜DF (x, θ) ≥ R if and only if CDF (x, θ, dc ) ≥ R. Equivalently, n o GDF (dc ) = (x cos θ, x sin θ) : C˜DF (x, θ) ≥ R + 2) CCF (x, θ) upper bounds CCF (x, θ, d) for all d ≥ dc , and therefore, [  + GCF (d) ⊂ (x cos θ, x sin θ) : CCF (x, θ) ≥ R d≥dc

Proof: Part 1 follows from the observation that when d = dc , by (1), any choice of ρ in (21) other than ρ = 0 would render CDF < R.

Proof: Observe, by (24) and (25) that C˜DF (x, θ) ≥ + CCF (x, θ) if and only if fmin ≥ d23,c2 where fmin is given by, fmin = inf (P2 d2 + d223 ) d≥dc

(27)

Let f (d) be defined by, f (d)

=

∆

P2 d2 + d223

=

(P2 + 1)d2 + x2 − 2dxs

(28)

where the last equality was obtained by the cosine theorem. Taking the derivative of f (d), we obtain that the unconstrained minimum (i.e., the minimum in the range d ∈ (−∞, ∞)) is obtained at, xs dmin = (29) P2 + 1 If s ≤ 0 then dmin ≤ 0 < dc , and thus fmin , which is the minimum in the range d ≥ dc , is obtained at dc . Thus, fmin = f (dc ) = P2 d2c + d223,c > d223,c , and our desired result of Part 1 of the lemma follows. To prove Part 2, we begin by considering the range x ∈ [0, x1 ]. We further divide the interval x ∈ [0, x1 ] into two overlapping intervals, [0, x0 ] and [x00 , x1 ] and prove the desired result separately in each interval. We define, 1 x0 = · dc s √ 1 P2 + 1 00 √ √ · dc x = s P2 + 1 + P2 Clearly, x00 < x0 < x1 , and thus the two intervals overlap. If x ≤ x0 then dmin < dc (by (29)), and thus fmin ≥ d223,c as in the proof of Part 1 of the lemma. Thus, the desired result is obtained for x ∈ [0, x0 ].

We proceed to consider the interval x ∈ [x00 , x1 ]. The minimum fmin (constrained to d ≥ dc ) clearly satisfies fmin ≥ f (dmin ). By (28) and (29), we have, after some algebraic manipulations,   s2 2 (30) f (dmin ) = x 1 − P2 + 1 Thus, a sufficient condition for fmin ≥ d223,c is,   s2 2 ≥ d223,c x 1− P2 + 1

(31)

Applying the cosine theorem and rearranging the terms, we obtain the inequality, s2 · x2 − (2dc s)x + d2c ≤ 0 (32) P2 + 1 The left hand side of this inequality is a parabola in x, whose roots are x00 and x1 . Thus, the inequality is satisfied for all x ∈ [x00 , x1 ]. Consequently, by the above discussion, fmin ≥ d223,c in this interval, as desired. This completes the proof on the lemma in the interval x ∈ [0, x1 ]. Before proceeding to the interval [x1 , ∞), observe that x1 satisfies (32) and consequently (31), with equality. Equivalently,   s2 2 = d223,c (33) x1 1 − P2 + 1 We now consider x ≥ x1 . Observe that the discussion leading to (30) remains valid in this range. By (25),   1 P2 +  CCF (x, θ) = log 1 + 2 + x inf (f (d, x, θ) d>dc   P2 1  ≤ log 1 + 2 + x inf f (d, x, θ) d>0   1 P (a)  2  = log 1 + 2 + s2 x 2 x 1 − P2 +1   (b) 1 P  2  ≤ log 1 + 2 + 2 x1 2 x1 1 − P2s+1 ! 1 P2 (c) = log 1 + 2 + 2 x1 d23,c =

C˜DF (x1 , θ)

where (a) is obtained by (30), (b) follows from x ≥ x1 , and (c) is obtained by (33). Lemmas 4 and 5 imply that a sufficient condition for Part 1 of Theorem 2 is C˜DF (x1 , θ) ≤ R for all θ such that s = cos θ > 0 (recall that x1 is a function of θ) and all P2 > 1/9 (recall that we have assumed that P1 = 1). Using (1) and (24), C˜DF (x1 , θ) ≤ R can be rewritten as, !   P2 1 1 log 1 + 2 + 2 ≤ log 1 + 2 (34) d23,c x1 dc

We now introduce the notations, √   P2 + 1 1 s2 √ , B = 1− A= √ s P2 + 1 − P2 P2 + 1

(35)

Observe that A and B are functions of s. With these notations, using (26) and (33), we obtain, x1 = A · dc and d223,c = Bx21 = BA2 · d2c . Plugging these identities into (34), with some algebraic manipulations, leads to the inequality, P2 ≤ B(A2 − 1)

(36)

The right hand side of the above inequality is a descending function of s in the range s ∈ [0, 1]. Thus, a necessary and sufficient condition for the inequality to be valid for all s ∈ [0, 1] is that the inequalities hold for s = 1. We thus confine our attention to this case. Using (35), we may rewrite (36) as, # √  " 2  P2 + 1 1 √ √ −1 P2 ≤ 1 − P2 + 1 P2 + 1 − P2 We now proceed to develop this inequality,   P2 P2 + 1 √ √ 2 −1 P2 ≤ P2 + 1 ( P2 + 1 − P2 ) p p (a) P2 + 1 ( P2 + 1 − P2 ) 2 ≤ P2 + 2 √ p p (b) P2 + 1 P2 + 1 − P 2 ≤ √ P2 + 2 p p p (c) (P2 + 1)(P2 + 2) ≤ P2 (P2 + 2) + P2 + 1 (d)

(P2 + 1)(P2 + 2)

≤

1 1 4

(e)

≤ ≤

P2 (P2 + 2) + P2 + 1 + p +2 P2 (P2 + 2)(P2 + 1) p 2 P2 (P2 + 2)(P2 + 1) P2 (P2 + 2)(P2 + 1)

(37)

(a) √ is obtained √ by multiplying both sides by (P2 + 1)/P2 · ( P2 + 1 − P2 )2 . (b) is obtained by taking the square root of √ both sides. (c) is obtained by multiplying both sides by P2 + 2 and rearranging terms. (d) is obtained by taking the squares of both sides. (e) is obtained by dividing by 2 and taking the square of both sides. Finally, the right hand side of (37) is an ascending function of P2 . The inequality can be verified to be satisfied for P2 = 1/9, and thus it is satisfied for all P2 > 1/9. B. Proof of Part 2 of Theorem 2 We begin with an outline of the proof. The proof follows from the observation that regardless of how low the power at the relay may be, if the destination is close enough to the relay, then arbitrarily high reliability may be achieved in the link between them. With CF, the relay and the destination cooperate and form a virtual antenna array. With DF, the relay must decode alone, and is thus confined to the no-relay coverage region (i.e., d must not exceed dc ). If its power is sufficiently low, it cannot service a destination that is beyond the no-relay region.

We confine our attention in this section to destinations a2 of region. We will later show that this implies that interior of A the form (0, x) where x > dc . We begin by examining CDF . is contained in the coverage region as well. By (21) (assuming, as discussed above, that P1 = 1), We must therefore show that at each point a3 on the ! √ boundary of A, P2 2ρ P2 1 lim sup CDF ≤ lim max log 1 + α + α + α/2 CDF (P2 = P1 , α = 2) ≥ R (39) P →0 0≤ρ≤1 x d 2 P2 →0 xα/2 d23 23     1 1 where CDF is given by (12). = log 1 + α < log 1 + α = R Note that by (38), at a point on the boundary of A, the x dc following holds, The last equality follows by (1). Thus, at any x > dc , for P2   that is too small, CDF < R, and thus we may not support the 1 √ √ 2 2 2 2 + λ cos(θ)) + ( γ sin(θ)) (40) d = d ( 13 desired rate of R. 2 We now turn to examine CCF . We wish to find a destination   location a3 = (0, x), where x > dc , such that regardless of 1 √ √ 2 2 2 2 how low the relay power P2 may be, by positioning the relay d23 = d ( − λ cos(θ)) + ( γ sin(θ)) (41) 2 close enough to the destination, we will be able to support the desired rate of R. Combining this with our above result for Consider (12). It is easy to observe that whenever CDF is greater or equal to R, then selecting ρ to coincide with (3) DF will conclude the proof. Let x > dc be arbitrary. We will examine relay locations would at worst reduce CDF to equal R. This is because this 1/α given by (0, d), where d = x − P2 . With this choice, choice renders the first minimization term equal to R and can only increase the second term. Thus, we assume, without loss P2 /dα 23 = 1. ! of generality, that this value of ρ maximizes (12). 1 1 Thus, proving (39) is equivalent to proving (recall that P1 = + lim CCF = lim log 1 + ˆ 2 ) xα P2 →0 P2 →0 dα (1 + N P2 ),     P1 2ρP1 P1 P1 Where we have used d12 = d, and d13 = x. Now, it is ≥ R = log 1 + 2 (1 − ρ2 ) . log 1 + 2 + 2 + straightforward to verify that the following holds, recalling d13 d23 d13 d23 d12 that d is now a function of P2 , and it satisfies P2 /dα 23 = 1. Substituting (40) and (41) into the above equations, after some ˆ2 ) = 2(1 + xα ) manipulations, we get (42) at the top of the page. lim dα (1 + N ∆ P2 →0 We now define s = sin2 (θ). We also define, Thus, 1 ∆   t(s) = + λ − (λ − γ)s (43) 1 1 4 + α lim CCF = log 1 + α ∆ P2 →0 2(1 + x ) x f (s) = t2 (s) − λ + λs (44) Now, taking the limit as x → dc , we obtain: Using this, equation (42) reduces to     1 1 + lim CCF lim = log 1 + 2t(s) 2ρ x→dc P2 →0 ) dα 2(1 + dα +p − (1 − ρ2 ) ≥ 0. c   c f (s) f (s) 1 > log 1 + α = R We will soon show (Lemma 6) that f (s) > 0, and thus, dc where the last equality follows by (1). Thus, there exists x0 > multiplying above equation by −f (s), we obtain that it is dc , such that for arbitrarily small P2 , the achievable rate with enough to show that p ∆ CF (when the relay is appropriately placed) is greater than R. h(s) = − 2t(s) − 2ρ f (s) + (1 − ρ2 )f (s) ≤ 0, This is exactly what we set out to prove, thus concluding the for all s ∈ [0, 1]. We will soon show (Lemma 7) that h(s) is proof of the theorem. zero on the end points of the interval s ∈ [0, 1]; hence it is A PPENDIX D enough to prove that h(s) is convex. This will soon be proved P ROOF OF T HEOREM 3 in Lemma 8. A. The Lower Bound The above discussion implies that the boundary of A lies within GDF . We now argue that this implies that the interior As noted in Section V, the proof of the lower bound follows of A is contained in GDF as well. Consider a point a3 in the by bounding the coverage region from within by an ellipse interior of A, and consider its polar coordinates (15). If x ≤ d, whose boundary is given by the points,   then it can easily be observed from (13) that CDF ≥ R (recall d √ √ that d = x, d12 = d and that we have assumed d ≤ dc ). Now, + λd cos(θ), γd sin(θ) (38) 13 2 keeping θ fixed, consider the range {x : d < x < xA (θ)}, We let A denote this ellipse. We begin by focusing on the where xA (θ) lies on the boundary of A. Both d13 and d23 boundary of A, and show that it lies within the DF coverage can easily be shown to be ascending functions of x in this

1 1 + 1 √ √ √ 2 2 + λ cos(θ)) + ( γ sin(θ)) ( 2 − λ cos(θ))2 + ( γ sin(θ))2 2ρ 1 q q ≥ 1 − ρ2 √ √ √ √ 1 1 2 2 2 2 (( 2 + λ cos(θ)) + ( γ sin(θ)) ) (( 2 − λ cos(θ)) + ( γ sin(θ)) ) ( 12

+

√

range, and thus by (12), CDF is a descending function of x. Therefore, if CDF ≥ R at the boundary point xA (θ), then it is greater than R at any point satisfying x < xA (θ). We now prove some of the lemmas that were used in the above discussion. Lemma 6: 1) λ − γ ≥ 21 0 2) t (s) < 0 for all s ∈ [0, 1] 3) f (s) is convex 4) f 0 (s) ≤ 0 for all s ∈ [0, 1] 4 5) f (s) ≥ (1−ρ) 2 > 0 for all s ∈ [0, 1] Proof:

1) p = −2t(0) − 2ρ f (0) + (1 − ρ2 )f (0)     1 1 = −2 + λ − 2ρ λ − 4 4  2 1 +(1 − ρ2 ) λ − 4 ! p 2(1 + ρ) 1 1 = −2 + + 2 1−ρ 1 − ρ2 ! p 2(1 + ρ) 1 + −2ρ 1−ρ 1 − ρ2 !2 p 2(1 + ρ) 1 2 +(1 − ρ ) + 1−ρ 1 − ρ2 ! p 2(1 + ρ) 1 1+ρ = −2 + + 2 1−ρ 1−ρ ! p 2(1 + ρ) 1+ρ 2 + 1+ +2 1−ρ 1+ρ 1+ρ

h(0)

1) 1 λ−γ− 2

= = =

p 2(1 + ρ) 1 1 2 1 1 + + − + − 4 1−ρ 1 − ρ2 1−ρ 4 2 p 2(1 + ρ) − (1 + ρ) 1 − ρ2 √ p 1+ρ √ ( 2 − 1 + ρ) ≥ 0 2 1−ρ

2) t0 (s) = −(λ − γ) < 0 by Part (1) of this Lemma. 3) f 00 (s) = 2t02 (s) = 2(λ − γ)2 > 0 by Part (1) of this Lemma. 4) f 0 (s) = 2t(s)(γ − λ) + λ. As t(s) is decreasing function (part (2) of this lemma), we need to prove that 2t(1)(γ − λ) + λ ≤ 0, or 2( 14 + γ)(λ − γ) ≥ λ. Or, ! p   2(1 + ρ) 1 2 1 − + 2 1−ρ 2 1−ρ 1 − ρ2 p 2(1 + ρ) 1 1 ≥ + + 4 1−ρ 1 − ρ2 After some manipulations, this is equivalent to proving r  2 1−ρ (3 + ρ) + ≥4 1+ρ 4 which is true since the left hand side decreases with ρ ∈ [0, 1) and goes to 4 as ρ → 1. 5) As f 0 (s) ≤ 0 (by previous part of this Lemma), f (s) ≥ 4 f (1) = t2 (1) = (1−ρ) 2. Lemma 7: h(s) = 0 on the end points of the interval s ∈ [0, 1], that is 1) h(0) = 0 2) h(1) = 0 Proof:

(42)

=

−1 −

=

0

2 1+ρ + 1−ρ 1−ρ

2) h(1)

p = −2t(1) − 2ρ f (1) + (1 − ρ2 )f (1)     1 1 + γ − 2ρ +γ = −2 4 4  2 1 +(1 − ρ2 ) +γ 4   2 2 ) = (1/4 + γ)(−2 − 2ρ + (1 − ρ ) 1−ρ = 0

Lemma 8: h(s) is convex. Proof: ∆

p

f (s) + (1 − ρ2 )f (s). p Since t(s) is linear, it is enough to show that −2ρ f (s) + (1 − ρ2 )f (s) is convex. Let √ ∆ g(x) = − 2ρ x + (1 − ρ2 )x, h(s) = − 2t(s) − 2ρ

for x > 0, we need to show g(f (s)) is convex. Since f (s) is convex (Lemma 6), g(x) is convex in the range {x : x > 0} (this is easily obeerved by, g 00 (x) = ρ2 x−3/2 > 0), it is enough

to show that g(x) is non-decreasing in its domain, and by a well-known convex analysis result (see, e.g. [11][Page 85]) we will obtain that h(s) is convex. To show that g(x) is non-decreasing in its domain, we see   2

ρ 4 . Since f (x) ≥ (1−ρ) that g 0 (x) ≥ 0 when x ≥ 1−ρ 2 ≥ 2  ρ by Lemma 6, g(x) is non-decreasing in the range of 1−ρ f (x).

We will soon show (Lemma 9) that f (s) > 0, and thus, multiplying above equation by f (s)/2, we obtain that it is enough to show that p 1 − ρ2 ∆ f (s) ≤ 0 h(s) = t(s) + ρ f (s) − 2

B. The Upper Bound As noted in Section V, the proof of the upper bound follows by bounding the coverage region from outside by a conic whose boundary is given by the points,  q d √ + λd cos(θ) 1 − a sin2 (θ), 2 q  2 1 − a sin (θ) √  (45) √ γd sin(θ) 1−a We let B denote this conic. The proof follows along lines similar to those of Appendix D-A. Once again, we begin by focusing on the boundary of the conic B, and show that on every point on this, CDF ≤ R. We will later show that this implies that exterior of the conic is not contained in the coverage region either. We must therefore show that at each point a3 on the boundary of B, CDF (P2 = P1 , α = 2) ≤ R

∆

f (s) = t2 (s) − λ(1 − s)(1 − as). With this notation, (49) becomes, 2t(s) 2ρ − (1 − ρ2 ) ≤ 0. +p f (s) f (s)

(46)

where CDF is given by (12). Note that at a point (45) the boundary of B, the following holds, " 2 q 1 √ 2 2 + λ cos(θ) 1 − a sin2 (θ) d13 = d 2 q  2  2 1 − a sin (θ) √   √ +  γ sin(θ) (47)  1−a

for all s ∈ [0, 1]. We will soon show (Lemma 10) that h(s) = 0 on the end points of the interval s ∈ [0, 1]; hence it is enough to prove that h(s) is convex which will follow from Lemma 10. The above discussion implies that the boundary of B, CDF ≤ R, with CDF = R only when sin2 θ = 0 or 1. We now argue that this implies that the exterior of B lies outside GDF . The discussion follows closely in the lines of a similar discussion in Appendix D-A. Consider an arbitrary point a3 , and consider its polar coordinates (15). We also let xB (θ) denote the value of x that coincides with the boundary of B (45). For all, x ≤ d, we have, as in Appendix D-A, CDF ≥ R. For θ such that s = sin2 θ ∈ / {0, 1}, we have shown above that CDF < R, and so xB (θ) > d. At the other values of θ, this can be shown by direct analytic calculation using (45). In the range x > d, CDF can be shown to be strictly descending (as in Appendix D-A). Using our results for the boundary of B, namely that CDF ≤ R when x = xB (θ), we obtain that for all x > xB (θ), CDF < R, and thus all such points (which constitute the exterior of B) lie outside of GDF . We now prove some results that were required in the above discussion. Lemma 9: 1) λ − γ ≤ 1 2) f 00 (s) ≤ 0 3) f 0 (s) ≤ 0 4) f (s) > 0 Proof: 1)

λ−γ−1 p 2(1 + ρ) 1 1 2 1 2 d23 = = + + + −1 − 4 1−ρ 1 − ρ2 1−ρ 4 p 2 2(1 + ρ) − 2(1 + ρ) − 1 + ρ2 = 2(1 − ρ2 ) p 2 2(1 + ρ) − 3 − 2ρ + ρ2 = 2(1 − ρ2 ) As in Appendix D-A, we may assume without loss of p It is easy to see that 2 2(1 + ρ) − 3 − 2ρ + ρ2 ≤ 0 generality that ρ in (12) is given by (3). With this choice, by proving convexity and checking the boundary points (46) becomes (recalling P1 = P2 ),     which completes the proof. P1 P1 2ρP1 P1 2 2) f 00 (s) = 2(aλ − 1)aλ = 2(λ − γ − 1)aλ ≤ 0 by first log 1 + 2 + 2 + ≤ R = log 1 + 2 (1 − ρ ) . d13 d23 d13 d23 d12 part of this Lemma. Substituting (47) and (48) into this equation, we get, after 3) f 0 (s) = 2t(s)t0 (s)+λ(1−as)+aλ(1−s). Since f 00 (s) ≤ some manipulations, (46) becomes (49) at the top of the 0 by the previous part of this Lemma, it is enough
to ∆ ∆ γ page. We now define s = sin2 (θ), t(s) = 14 + λ(1 − as) and prove f 0 (0) ≤ 0. Or, f 0 (0) = λ 23 − 2λ − 2(λ − γ) ≤ "

2 q 1 √ 2 2 d − λ cos(θ) 1 − a sin (θ) 2 q  2  2 1 − a sin (θ) √   √ +  γ sin(θ) (48)  1−a

1 4

+ λ(1 − a sin2 (θ)) +

r 1 4

1 q

+ λ(1 − a sin2 (θ)) cos(θ)

1 4

+ λ(1 − a sin2 (θ)) −

1 q

+

(49)

λ(1 − a sin2 (θ)) cos(θ)

2ρ r ≤ 1 − ρ2 q q + λ(1 − a sin2 (θ)) + λ(1 − a sin2 (θ)) cos(θ) 14 + λ(1 − a sin2 (θ)) − λ(1 − a sin2 (θ)) cos(θ)

0. Hence, it is enough to prove that 3λ − γ ≤ 4λ(λ − γ). After substituting the values of λ and γ and some 2 2 − (1−ρ) ≥ manipulations, it reduces to proving that 1+ρ 8 1 which is true since left hand side decreases with ρ ∈ [0, 1) from 15/8 at ρ = 0 to 1 as ρ → 1. 4) As f 0 (s) ≤ 0 by the previous part of this Lemma, it is enough to prove f (1) > 0.

f (1) = t2 (1) =



2t(1) 2ρ − (1 − ρ2 ) +p f (1) f (1) 2t(1) 2ρ = + − (1 − ρ2 ) 2 t (1) t(1) 1+ρ = 2 − (1 − ρ2 ) t(1) 1+ρ − (1 − ρ2 ) = 2 1/4 + γ 1+ρ = 2 2 − (1 − ρ2 )

2  2 1 1 > 0. + λ(1 − a) = +γ 4 4

1−ρ

= Lemma 10: 1) h(0) = 0 2) h(1) = 0 3) h000 (s) ≤ 0 4) h00 (s) ≥ 0

3) p 1 − ρ2 h(s) = t(s) + ρ f (s) − f (s) 2

Proof: 1) Proving this is equivalent to prove

2t(0) f (0)

+ √2ρ

f (0)

− (1 −

2

ρ )=0

= = = = =

= =

2) Proving this is equivalent to proving, (1 − ρ ) = 0

1 − ρ2 0 ρf 0 (s) − f (s) h0 (s) = t0 (s) + p 2 2 f (s) h00 (s)

2t(0) 2ρ +p − (1 − ρ2 ) f (0) f (0) 2(λ + 1/4) 2ρ + − (1 − ρ2 ) 2 (λ − 1/4) λ − 1/4 2(λ − 1/4) + 1 2ρ p − (1 − ρ2 )(1 − ) (λ − 1/4)2 1 + ρ + 2(1 + ρ) p 2 1 1 + − (1 − ρ + 2(1 + ρ)) 2 λ − 1/4 (λ − 1/4) λ − 1/4   p 1 (1 + ρ − 2(1 + ρ))(λ − 1/4) + 1 (λ − 1/4)2 p 1 ((1 + ρ − 2(1 + ρ)) 4) 2 (λ − 1/4) p (1 + ρ + 2(1 + ρ)) + 1) 1 − ρ2 1 (−1 + 1) (λ − 1/4)2 0

2

0

2t(1) f (1)

+ √2ρ

f (1)

−

ρ = t00 (s) + 2(f (s))(3/2)   1 − ρ2 00 (f 0 (s))2 f (s) − f (s)f 00 (s) − 2 2

As t(s) is linear, h00 (s)

=

h000 (s) =

  ρ (f 0 (s))2 00 f (s)f (s) − 2 2(f (s))(3/2) 1 − ρ2 00 − f (s) 2

−3ρ f 0 (s)(f (s)f 00 (s) − f 0 (s)2 /2) 4f (s)5/2

By Lemma 9, we see that h000 (s) ≤ 0 00 Using h000 (s) ≤ 0, it is enough to show that q  h (1) ≥ 0. 1 4 2 0 f (1) = − 4 − (1−ρ)2 f 00 (1) = 1+ρ − 1   √ 2(1+ρ) 1 2 2 − 41 + (1−ρ) + − 2 2 (1−ρ2 )(1−ρ) (1−ρ2 )(1−ρ)
1−ρ 00 00 h (1) = −f (1)(2 + ρ) − ρ8 (1 − ρ)2 f 0 (s)2 4 After some manipulations, proving h00 (1) ≥ p0 is equivalent to proving (1 − ρ)2 − 4(1 + ρ) − 8 + 8 q 2(1 + ρ) + ρ ρ 3 2 2 4 2 4 ρ(1 − ρ) − 128 (1 − ρ) − 4 (1 − ρ) 1+ρ ≥ 0 Hence, we will be done if we prove p a) (1 − ρ)2 − 4(1 + ρ) − 8 + 8 2(1 + ρ) ≥ 0

b)

3 4 ρ(1

− ρ)2 −

ρ 128 (1

− ρ)4 − ρ4 (1 − ρ)2

q

2 1+ρ

≥0

which are true since 1) is monotonically increasing with ρ and 2) is monotonically decreasing with ρ, and the values at boundary points are satisfied.

Proof: 1) h(0)

=

A PPENDIX E P ROOF OF T HEOREM 4 The proof follows closely in the lines of Appendix D-A. We review its main points below. As in that appendix, this proof follows by bounding the coverage region from within by an ellipse whose boundary is given by the points (38), but with λ and γ as defined by (9) and (10).We once again let A denote this ellipse, and begin by focusing on the boundary of A, and show that it lies within the DF coverage region. By a similar discussion to that provided in Appendix D-A, this will imply that the interior of A is contained in the coverage region as well. We must therefore show that at each point a3 on the boundary of A, CDF (P2 = P1 , α = 4) ≥ R

f 2 (0) 2λ − f (0) − 2 1+ρ (t2 (0) − λ)2 2λ (1 − ρ) − (t2 (0) − λ) − 2 1+ρ (λ − 14 )4 2(λ − 14 ) 1 1 (1 − ρ) − (λ − )2 − − 2 4 1+ρ 2(1 + ρ) 1−ρ 1 2 1 [(λ − )4 − (λ − )2 − 2 4 1−ρ 4 4 1 1 (λ − ) − ] 2 1−ρ 4 (1 − ρ2 ) 1−ρ [0] 2 0

(1 − ρ)

=

= =

= = 2)

h(1)

(50)

2(f (s) + 2λ(1 − s)) 2ρ + − (1 − ρ2 ) ≥ 0. f 2 (s) f (s) We will soon show (Lemma 12) that f (s) > 0, and thus, f 2 (s) multiplying above equation by − 2(1+ρ) , we obtain, ∆

h(s) = (1 − ρ)

f 2 (s) 2λ(1 − s) − f (s) − ≤0 2 1+ρ

for all s ∈ [0, 1]. We will soon show (Lemma 11) that h(s) is zero on the end points of the interval s ∈ [0, 1]; hence it is enough to prove that h(s) is convex. This will soon be proved in Lemma 13. The above discussion implies that the boundary of A lies within GDF . As in Appendix D-A, this implies that the interior lies within GDF as well. We now prove some results that were required in the above discussion. Lemma 11: 1) h(0) = 0 2) h(1) = 0

=

0

=

As in Appendix D-A, we begin by observing Note that by (38), at a point on the boundary of A, d13 and d13 satisfy (40) and (41). Once again, as in Appendices D-A and D-B, we may assume without loss of generality that ρ in (12) is given by (8). With this choice, (50) becomes (recalling P1 = P2 ),     P1 2ρP1 P1 P1 2 ≥ R = log 1 + 4 (1 − ρ ) . log 1 + 4 + 4 + 2 2 d13 d23 d13 d23 d12 Substituting (40) and (41) into this equation, after some manipulations, we obtain (51) at the top of the page. We define s, t(s) and f (s) as in Appendix D-A (see (43) and (44)). Using this notation, (51) becomes,

=

f 2 (1) − f (1) 2 t4 (1) − t2 (1) (1 − ρ) 2 1−ρ 2 t2 (1)(1 − t (1)) 2 1−ρ 1 t2 (1)[1 − (γ + )2 ] 2 4 1−ρ 2 2 ] t (1)[1 − 2 1−ρ t2 (1)[1 − 1]

=

= = =

(1 − ρ)

Lemma 12: 1) f (s) is convex 2) f 0 (s) ≤ 0 for all s ∈ [0, 1] 2 3) f (s) ≥ 1−ρ >0 Proof: 1) f 00 (s) = 2t02 (s) = 2(λ − γ)2 ≥ 0. 2) As f 00 (s) ≥ 0, it is enough to prove f 0 (1) ≤ 0. f 0 (1)

2t(1)t0 (1) − λ(−1) 1 = −2( + γ)(λ − γ) + λ 4 r r 2 2 1 = −2 (λ − + )+λ 1−ρ 1−ρ 4 q 2 To prove f 0 (1) ≤ 0, we need to prove −2 1−ρ (λ − q 2 1 1−ρ + 4 ) + λ ≤ 0, or =

p 1 1 1 √ 2 >√ [ (2 2 − 1 − ρ) + √ ] √ 4 1−ρ 4 2 2− 1−ρ √ √ Let d = 14 (2 2 − 1 − ρ) + 2√2−2√1−ρ . It is enough to show that d 2 d 4 d 1 √ (√ )4 − (√ )2 − − ≤0 2 1−ρ 1−ρ 1−ρ 1−ρ 1 − ρ 1 − ρ2 λ−

1

+



1 4



1 4

+ λ cos2 θ + γ sin2 θ − 2

+ λ cos2 θ + γ sin θ −

√ √

λ cos θ λ cos θ

2 +  2ρ 

√ √ Let q = 2 2 √ − 1 − ρ. It is enough to prove d2 (d2 − 2)(1+ρ)−4d 1 − ρ−(1−ρ) ≤ 0. It is enough to prove √ q2 )( 2q − 4q )(1+ d(d2 −2)(1+ρ)−4 1 − ρ ≤ 0, or ( q42 − 16 √ ρ) 6 4 1 − ρ. Hence, we will be done if we prove 2

q a) q42 − 16 ≤1 √ q 2 b) q − 4 ≤ 2 1 − ρ c) 1 + ρ ≤ 2 √ √ where (a) follows since q 2 ≥ (2 2 − 1)2 ≥ 8(√ 2 − 1), 4 2 Hence, √ q +16q −64 ≥ 0, (b) is equivalent to 1 − ρ ≤ 12 2 which is true and (c) follows since ρ ≤ 1. 7 3) As f 0 (s) ≤ 0 (by second part of this Lemma), f (s) ≥ 2 f (1) = t2 (1) = 1−ρ .

Lemma 13: h(s) is convex. Proof: ∆

h(s) = (1 − ρ)

2λ(1 − s) f 2 (s) − f (s) − 2 1+ρ

As 2λ(1−s) is linear, it is enough to show that (1 − ρ) f 1+ρ f (s) is convex. Let ∆

g(x) = (1 − ρ)

2

(s) 2

−

x2 − x, 2

we need to show g(f (s)) is convex. Since g(x) is convex (g 00 (x) = 1 − ρ > 0) and f (s) is convex (Lemma 12), it is enough to show that g(x) is non-decreasing in its domain, and as in the proof of Lemma 8, we will obtain that h(s) is convex. To show that g(x) is non-decreasing in in the range of f (x), 1 2 we see that g 0 (x) ≥ 0 when x ≥ 1−ρ . Since f (x) ≥ 1−ρ > 1 1−ρ by Lemma 12, g(x) is non-decreasing in the range of f (x). A PPENDIX F P ROOF OF L EMMA 1

π

p λγd2

p

λγ(1 − ρ2 )d2c q p p = πd2c (λ 1 − ρ)(γ 1 − ρ)(1 + ρ) (52) = π

We now variables from d to ρ, using (8). Taking d → 0 is equivalent to taking ρ → 1.

lim γ

ρ→1

p √ 1−ρ= 2

(53)

1 4

1 1 4

+ λ cos2 θ + γ sin2 θ +

+ λ cos2 θ + γ sin2 θ +

√

√

λ cos θ

λ cos θ

2

 ≥ 1 − ρ2

(51)

√ By definition of λ (Theorem 4), λ 1 − ρ is the largest solution of  4  2 t 1 2 t 1 √ √ − − − 1−ρ 1−ρ 4 1−ρ 4   1 4 t 1 √ − − =0 − 2 1−ρ 1 − ρ2 1−ρ 4 Multiplying by (1 − ρ)2 4  2  √ √ 1−ρ 1−ρ −2 t− − t− 4 4   √ √ 4 1−ρ 1−ρ 1−ρ =0 t− − 1+ρ 4 1+ρ √ √ Thus, λ 1 − ρ = 1 − ρ/4 + ψ, where ψ is the largest solution of √ 4 1−ρ 1−ρ t4 − 2t2 − t− =0 1+ρ 1+ρ It is easy to see that as √ ρ → 1, the largest solution of above equation approaches 2 (roots vary continuously with the coefficients of the polynomial, hence the limit of the root of the polynomial equals the root of the limiting polynomial [12]). Hence, √ p √ 1−ρ √ lim λ 1 − ρ = lim (54) + 2= 2 ρ→1 ρ→1 4 Substituting (53) and (54) in (52), we obtain, p lim π λγd2 = 2πd2c . d→0

We now consider the true coverage region when d → 0. Taking d → 0 is equivalent to placing the relay at the source. Examining (12), it is straightforward to observe that this √ produces a coverage region which is a circle with radius 2dc . Thus, the true area coincides with the above computed limit of the lower bound, when d → 0. A PPENDIX G P ROOF OF T HEOREM 5 The proof of this theorem follows along lines similar to those of Theorem 1. The half-duplex DF and CF achievable rates were computed by Zhixin et al. [10], and are given by,    P1 CDF = max min t log 1 + α + (1 − t)× 0≤ρ≤1 d   12   P P1 1 2 log 1 + (1 − ρ ) α , t log 1 + α + d13 d √ 13   P1 P2 2ρ P1 P2 (1 − t) log 1 + α + α + d13 d23 (d13 d23 )α/2 (55)

CCF

=

P1 P1 + α α ˆ2 ) d13 d12 (1 + N   P1 +(1 − t) log 1 + α d13

!

t log 1 +

(56)

where ˆ2 = N

1 + P1 ( d1α + d1α ) 12 13   P2 /dα 23 (1−t)/t − 1 (1 + dPα1 ) (1 + 1+P ) α 1 /d 13

(57)

13

We begin by proving GCF (d) ⊇ GNR (d), for all d > 0. This is straightforward from the observation that by (56) and (16), CCF > CNR at every destination point a3 . We now prove Part 2 of the theorem, which (as in the proof of Theorem 1) is the easier of the two parts. To show that GNR > GDF (d), we first note that by (16) and (1), it can easily be observed that GNR is exactly a sphere with radius dc . We now show that DF cannot support a rate of R outside this sphere. Let θ ∈ [0, 2π), x > dc , and consider the achievable rate CDF at the point a3 whose polar coordinates (see Appendix A-B) are (θ, x),     (a) P1 P1 CDF ≤ t log 1 + α + (1 − t) log 1 + α d12 d13     (b) P1 P1 < t log 1 + α + (1 − t) log 1 + α dc dc   P1 = log 1 + α dc (c)

=

R

(58)

where (a) follows from (55), (b) follows since x > dc , d > dc , and (c) follows from (1). Thus, CDF is strictly less that R outside the sphere of radius dc , as desired. To prove the first part of the theorem, we again apply the notation xS (θ) that was defined in Appendix B. By (55), we may obtain a lower bound on CDF by restricting the maximization to ρ = 0. Thus (recall that we have assumed t = 1/2),      P1 P1 1 CDF ≥ min log 1 + α + log 1 + α , 2 d12 d13     P1 P1 P2 log 1 + α + log 1 + α + α d13 d d    13  23  1 P1 1 P1 = log 1 + α + min log 1 + α , 2 d13 2 d12   P1 P2 log 1 + α + α (59) d13 d23 The following lemma examines this expression at x = xDF (θ). Lemma 14: For all θ, consider the point a3 whose polar coordinates are given by (θ, xDF (θ)). Then (18) holds at a3 , Proof: Assume, by contradiction, that (18) does not hold. Letting x = xDF (θ), we will now show (as in the proof of

Lemma 3) that we may increase x and preserve CDF ≥ R, contradicting the maximality of xDF (θ). Consider (55). Whenever (18) does not hold, we have by (59),       1 1 1 P1 P1 P1 log 1 + α + log 1 + α > log 1 + α CDF ≥ 2 d13 2 d12 2 d12   (a) 1 P1 (b) ≥ log 1 + 0α = R 2 dc Where the (a) was obtained by the fact that d12 = d and d ≤ d0c , and (b) was obtained by (11). Since CDF > R, by a continuity argument, we may increase x slightly and still get CDF ≥ R. Therefore, if CDF > R at x = xDF (θ), it will remain so after we increase x a little, producing the desired contradiction. By Lemma 14, (17) implies that at x = xDF (θ),     1 P1 1 P1 P2 CDF ≥ log 1 + α + log 1 + α + α 2 d13 2 d13 d23 > CCF The last inequality is obtained by simple arithmetic, using (56) (recalling that t = 1/2) in a similar way to the way (20) was obtained in Appendix B. Again, by similar arguments as in Appendix B, this implies that CCF < R for all x > xDF (θ), and thus GDF (d) ⊃ GCF (d) as desired. This completes the proof of Part 1 of the theorem. A PPENDIX H E XTENSIONS TO P HASE FADING A. Extension of Theorem 1 to Phase Fading The achievable rates with DF and CF and the upper bound, which were computed in [5] are given by,    P1 CDF = min log 1 + α , d12   P1 P2 log 1 + α + α (60) d13 d23 CCF is as given in Appendix A, expressions (13) and (14).    1 1 CUB = min log 1 + P1 ( α + α ) , d12 d   13 P1 P2 (61) log 1 + α + α d13 d23 The proof follows closely in the lines of Appendix B. The expressions for CCF and CNR remain unchanged in the phase fading setting and thus the result GCF (d) ⊇ GNR (d) carries over immediately. Examining (60), it is easily observed to be equal to (12) with the maximization replaced with an assignment of ρ = 0. The remainder of the proof follows along direct lines as Appendix B and is omitted.

B. Proof of Lemma 2 Lemma 3 (Appendix B) can straightforwardly be shown to carry over to the phase-fading setting. With this lemma, at the point whose polar coordinates are (θ, xDF (θ)) (for arbitrary α θ), (61) reduces to CUB = log (1 + P1 /dα 13 + P2 /d23 ), which coincides with CDF at that point. By continuity arguments, CDF must equal R at (θ, xDF (θ)) and thus CUB equals R as well. In Appendix B we showed that the value of CCF at a point whose polar coordinates are (θ, x), descends as a function of x for x > xDF (θ). Similar arguments can be now be applied to CUB , again rendering GUB (d) ⊆ GDF (d). Since we trivially have GUB (d) ⊇ GDF (d), this completes the proof of the lemma. R EFERENCES [1] E. C. van der Meulen, “Transmission of information in a T-terminal discrete memoryless channel,” Ph.D. dissertation, Univ. California, Berkeley, CA, Jun. 1968. [2] E. C. van der Meulen, “Three-terminal communication channels,” Adv. Appl. Probab., vol. 3, pp. 120-154, 1971. [3] E. C. van der Meulen, “A survey of multiway channels in information theory: 1961-1976,” IEEE Trans. Inf. Theory, vol. IT-23, no. 1, pp. 137, Jan. 1977. [4] T. M. Cover and A. A. El Gamal, “Capacity theorems for the relay channel,” IEEE Trans. Inf. Theory, vol. IT-25, no. 5, pp. 572-584, Sep. 1979. [5] G. Kramer, M. Gastpar, and P. Gupta, “Cooperative strategies and capacity theorems for relay networks,” IEEE Trans. Inf. Theory, vol. 51, no. 9, pp. 3037-3063, Sep. 2005. [6] J.-P. Tignol, “Galois theory of algebraic equations,” Longman Scientific and Technical Press, 1988. [7] J. N. Laneman, D. N. C. Tse and G. W. Wornell, “Cooperative Diversity in Wireless Networks: Efficient Protocols and Outage Behavior,” IEEE Trans. Inf. Theory, vol. 50, no. 12, pp. 3062-3080, Dec. 2004. [8] A. Høst-Madsen and J. Zhang, “Capacity bounds and power allocation for wireless relay channels,”IEEE Trans. Inf. Theory, vol. 51, no. 6, pp.2020- 2040, June 2005 [9] M. A. Khojastepour, A. Sabharwal, B. Aazhang, “On capacity of Gaussian ‘cheap’ relay channel,” Globecom 2003 [10] L. Zhixin, V. Stankovic and X. Zixiang, “Wyner-Ziv coding for the halfduplex relay channel,” ICASSP 2005 [11] S. Boyd and L. Vandenberghe, “Convex Optimization,” Cambridge University Press, 2004 [12] D. J. Uherka and A. M. Sergott, “On the Continuous Dependence of the Roots of a Polynomial on its Coefficients”, The American Mathematical Monthly, vol. 84, No. 5. pp. 368–370, May 1977.